Photodiode Current/Short Circuit Current

Thread Starter

Sparky49

Joined Jul 16, 2011
833
Hi all.

I was wondering if someone could clarify something regarding photodiodes. I'm trying to design a circuit to measure light intensity, using a photodiode. I'm using the VTB8440B (http://www.hofoo.com.cn/uploadfiles/VTB datasheet.pdf written page 42). The problem is I've seen some many different variations of the op amp circuit to increase the output that I'm confused as to what I actually need in my application.

I have a single +5V supply, and am using the TL972IP op amp (http://www.ti.com.cn/cn/lit/ds/symlink/tl972.pdf).

I've come across the following circuit. Strictly speaking, my understanding is Vout=-(Rf*Id), but as current flow is in the direction drawn, we will get a positive output voltage, correct?

DSC_0256.JPG

Now, the part I most confused about is determining Id. What determines Id? Is it simply a different term for Isc given in the datasheet, or something else? If I wish to determine the light intensity accurately (ie, get actual values instead of 'is brighter/darker than') is this the way to do it, or is another approach required?

Many thanks for all of your time. If something is not clear let me know.

Sparky
 
I've come across the following circuit. Strictly speaking, my understanding is Vout=-(Rf*Id), but as current flow is in the direction drawn, we will get a positive output voltage, correct?
You are correct.

The current that is flowing in the photodiode is what you are referring to as Id. For purposes of your circuit, that is the same as Isc that is listed on the datasheet. On page 42 of the datasheet the Isc is given for one test condition, that is, for one light intensity and temperature.

There is a graph on page 22 that shows the relationship of Isc and light intensity in foot-candles. The graph is titled "Relative Short Circuit Current vs. Illumination." These are not exact values but are relative to the Isc given on page 42. This can help you determine the intensity of light presented to the detector.

As a side note, Id is used for dark current on your datasheet. That is the current flowing through the diode when the light intensity is 0.
 

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Thread Starter

Sparky49

Joined Jul 16, 2011
833
Many thanks for the reply!

Ahhh, I presumed Id was diode current or something, oops! :p

Would you mind checking my calculations? I also went and checked the minimum detectable illumination, a very tiny current. Is such a small current detectable by the op amp?

DSC_0257.JPG DSC_0258.JPG

Many thanks again.

Sparky
 
The calculations look reasonable.

Such tiny currents may or may not be distinguishable in your system. The currents are also small enough to be on the same order of magnitude as the other non-ideal currents flowing through the operational amplifier itself, such as input bias currents and input offset currents. These can cause offsets in your readings that may need to be calibrated out.

There is such a small voltage being generated that the output may be sitting in system noise.

Depending on the final accuracy of the device you need, it can get a little tricky. You may have to set up some slightly different illumination ranges...
 

Thread Starter

Sparky49

Joined Jul 16, 2011
833
Thanks again for the reply.

Yeah I suspected the lower range might be too small to register. I was planning to at least two separate circuits to cover an order of magnitude of illumination or so each. I will solder up a couple tomorrow. Thanks again.
 

Alec_t

Joined Sep 17, 2013
10,746
Given that no two photo-detectors, even from the same batch, are likely to have a very close match in performance, how do you plan to calibrate your system?
 
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